Question

An object is thrown upward from the edge of a tall building with a velocity of 10 m/s. Where will the object be 3 s after it is thrown? Take g=10ms^2
A. 15m above the top of the building
B. 30 m below the top of the building
C. 15 m below the top of the building
D. 30 m above the building

Answer 1

use of uniform accelerated motion equations are required here.....can help u if u can say where u got stuck

Answer 2

okay, i tried calculating the distance to the top using S=ut+1/2gt^2 and got 45m but i can't really the fix the part of its coming down.

Answer 3

lets break up this question into parts....can u find the time when the stone reaches maximum height(when velocity=0)? part 1

Answer 4

using v=u + at with negative acceleration due to gravity and final velocity of 10m/s, i got 1s for time to the maximum height

Answer 5

yes!

Answer 6

then we go to part 2 where for the remaining 2 seconds i want the dispalcement WITH RESPECT to ground from that max height point where u=0 i have t and have a so find s

Answer 7

using S = ut + 1/2gt^2 i got 20m. right?

Answer 8

yes

Answer 9

thats right and that is the height from the ground(dispalcement with respect to ground level)

Answer 10

i still did not get it, do you have an idea on how to go about it?

Answer 11

u got the answer 20 m above the ground thats all.......wat dint u get?

Answer 12

the answer is 15m above the building

Answer 13

@woleraymond, you are right the answer is 15m above, it will be nice to show the work out.
and here it is:

first calculate the distance covered by the object

v^2 = u^2 + 2as where a = g = 10m/s^2 and since it is an upward motion g = -10m/s^2

therefore, 0 = 10^2 + 2(-10)s
0 = 100 - 20s and 20s = 100 therefore, s = 100/20 = 5m.

now, calculate the time taken to travel this distance
therefore, v = u + gt
0 = 10 + (-10)t -------> 0 = 10 - 10t ------> 10t = 10 and t = 10/10 = 1s

therefore, if it took 1s to travel 5m, the object will 15m in 3s. so the answer is 15m. I hope this will help someone.